 Lesson: Squares and Square Roots
Exercise: 6.1
Question 1
What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234
(vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555
(i) 1 (ii) 4 (iii) 1 (iv) 9 (v) 6 (vi) 9
(vii) 4 (viii) 0 (ix) 6 (x) 5
Solution:
(i) 1 (ii) 4 (iii) 1 (iv) 9 (v) 6
(vi) 9 (vii) 4 (viii) 0 (ix) 6 (x) 5
Question 2
The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000
(vi) 89722 (vii) 222000 (viii) 505050
(i) Numbers ending in 7 are not square numbers.
(ii) Numbers ending in 3 are not square numbers.
(iii) Numbers ending in 8 are not square numbers.
(iv) Numbers ending in 2 are not square numbers.
(v) Numbers ending in odd number of zeros are not square numbers.
(vi) Numbers ending in 2 are not square numbers.
(vii) Numbers ending in odd number of zeros are not square numbers.
(viii) Numbers ending in odd number of zeros are not square numbers.
Solution:
(i) Numbers ending in 7 are not square numbers.
(ii) Numbers ending in 3 are not square numbers.
(iii) Numbers ending in 8 are not square numbers.
(iv) Numbers ending in 2 are not square numbers.
(v) Numbers ending in odd number of zeros are not square numbers.
(vi) Numbers ending in 2 are not square numbers. (vii)Numbers ending in odd number of zeros are not square numbers.
(viii) Numbers ending in odd number of zeros are not square numbers.
Question 3
The square of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
(i) and (iii)
Solution:
(i) and (iii)
Question 4
Observe the following pattern and find the missing digits:
2
2
2
2
2
11 121
1 01 10201
1 001 1002001
1 00001 1......2.......1
1 0000001 .......................
10000200001, 100000020000001
Solution:
2
100001 10000200001
2
10000001 100000020000001
Question 5
Observe the following pattern and find the missing digits:
2
2
2
2
2
11 121
1 01 10201
1 0101 102030201
1 010101 ....................
...... 10203040504030201
1020304030201, 101010101
2 Solution:
2
1010101 1020304030201
2
101010101 10203040504030201
Question 6
Using the given pattern, find the missing numbers:
2 2 2 2
2 2 2 2
2 2 2 2
22
1 2 2 3
2 3 6 7
3 4 1 2 1 3
4 5
2
2 2 2
2 2 2 2
_2 21
5 _2 30 31
6 7 _ _

20, 6, 42, 43
Solution:
2 2 2 2
4 5 20 21
2 2 2 2
6 7 42 43
Question 7
(i)
1 3 5 7 9
(ii)
1 3 5 7 9 11 13 15 17 19
(iii)
1 3 5 7 9 11 13 15 17 19 21 23
(i) 25 (ii) 100 (iii) 144
Solution:
(i)
2
1 3 5 7 9 5
[Sum of first five odd natural numbers] = 25
(ii)
1 3 5 7 9 11 13 15 17 19
= (10)
2
[Sum of first ten odd natural numbers]
= 100
(iii)
1 3 5 7 9 11 13 15 17 19 21 23
= (12)
2
[Sum of first twelve odd natural numbers]
= 144 Question 8
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
(i)
1 3 5 7 9 11 13
(ii)
1 3 5 7 9 11 13 15 17 19 21
Solution:
(i)
2
49 7 Sum of first 7 odd numbers 1 3 5 7 9 11 13
2
(ii) 121 11 Sum of first 11 odd numbers 1 3 5 7 9 11 13 15 17 19 21
Question 9
How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
(i) 24 (ii) 50 (iii) 198
Solution:
(i) Here,
12n
and
1 13n 
2n non perfect square numbers lie between
2
2
& 1 .nn
Thus,
2 12 24
such numbers lie between 12
2
and 13
2
,
(ii) Similarly,
2 25 50
such numbers lie between 25
2
and 26
2
.
(iii) Similarly,
2 99 198
such numbers lie between 99
2
and 100
2
.
Exercise 6.2
Question 1
Find the square of the following numbers without actual multiplication:
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
(i) 1024 (ii) 1225 (iii) 7396 (iv) 8649 (v) 5041 (vi) 2116 Solution:
(i)
2
2
32 30 2 30 2 30 2 30 30 2 2 30 2
22
30 60 60 2 900 120 4 1024
(ii)
2
35 3 4 hundreds 25 12 100 25 1200 25 1225
(iii)
2
2
86 80 6 80 6 80 6 80 80 6 6 80 6
22
80 80 6 6 80 6 6400 480 480 36 7396
(iv)
2
2
93 90 3 90 3 90 3 90 90 3 3 90 3
2 2
90 90 3 3 90 3 8100 270 270 9 8649
(v)
2
2
71 70 1 70 1 70 1 70 70 1 1 70 1
2
70 70 1 1 70 12 4900 70 70 1 5041
(vi)
2
46 40 6 40 6 40 6 40 40 6 6 40 6
22
40 40 6 6 40 6
1600 240 240 36 2116
Question 2
Write a Pythagorean triplet whose one member is:
(i) 6 (ii) 14 (iii) 16 (iv) 18
(i) 6,8,10 (ii) 14, 48, 50 (iii) 16, 63, 65 (iv) 18, 80, 82
Solution:
We have the Pythagorean triplet
22
2 , 1, 1 ;m m m
where
1m
(natural number).
(i)
2 2
1 6 6 1 7Let m m
Or,
7m
(Not a natural number)
2
1 6m 
2
6 – 1 5m
Or,
5m
(not a natural number)
And
2 6 m
3m
(A natural number > 1) Thus,
22
– 1 3 – 1 9 – 1 8m
and
22
1 3 1 9 1 10m
Hence, the required triplet is 6, 8, and 10
(ii) Let
2
1 14 m
2
14 1 15m
Or,
15 m
(not a natural number)
2
1 14 m 
2
14 1 13m
Or,
13m
(not a natural number)
And
2 14 m
7 m
(A natural number> 1)
Thus,
22
1 7 1 49 1 48m
and
22
1 7 1 49 1 50m
Hence, the required triplet is 14, 48, and 50
(iii) Let
2
1 16 m
2
16 1 17m
Or,
17m
(not a natural number)
or,
2
1 16 m 
2
16 – 1 15 m
Or,
15m
(not a natural number)
And
2 16 m
8m
(A natural number> 1)
Thus,
22
1 8 – 1 64 – 1 63m
and
22
1 8 1 64 1 65m
Hence, the required triplet is 16, 63, and 65.
(iv) Let
22
1 18 18 1 19mm
Or,
19m
(not a natural number)
2
1 18 m 
2
18 – 1 17m
Or,
17m
(not a natural number) and
2 18 m
9 m
(A natural number> 1)
Thus,
22
1 9 1 81 1 80m
and
22
1 9 1, 81 1 82m
Hence, the required triplet is 18, 80, and 82. Exercise 6.3
Question 1
What could be the possible ‘one’s’ digits of the square roots of each of the
following numbers?
(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025
(i) 1, 9 (ii) 4, 6 (iii) 1, 9 (iv) 5
Solution:
(i) Unit’s digit of square root of 9801 is 1 or 9.
(ii) Unit’s digit of square root of 99856 is 4 or 6.
(iii) Unit’s digit of square root of 998001 is 1 or 9.
(iv) Unit’s digit of square root of 657666025 is 5.
Question 2
Without doing any calculation, find the numbers which are surely not perfect
squares.
(i) 153 (ii) 257 (iii) 408 (iv) 441
(i), (ii), (iii)
Solution:
(i) 153 (ii) 257 (iii) 408 are surely not perfect squares.
Because the numbers that end
with 2, 3, 7 or 8 can never be a perfect square.
Question 3
Find the square roots of 100 and 169 by the method of repeated subtraction.
10, 13
Solution:
For 100,
100 1 99
99 3 96
96 5 91 91 7 84
84 9 75
75 11 64
64 13 51
51 15 36
36 17 19
19 19 0
We observe that the number 100 got reduced to zero after subtracting 10
consecutive odd numbers starting from 1.
Thus, 100 is a perfect square.
2
100 10
2
100 10 10
For 169
169 1 168
168 3 165
165 5 160
160 7 153
153 9 144
144 11 133
133 13 120
120 15 105
105 17 88
88 19 69
69 21 18
48 23 25
25 25 0
We observe that the number 169 got reduced to zero after subtracting 13
consecutive odd numbers starting from 1.
Thus, 169 is a perfect square.
2
169 13
2
169 13 13 Question 4
Find the square roots of the following numbers by the Prime Factorization
Method:
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744
(vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100
(i) 27 (ii) 20 (iii) 42 (iv) 64 (v) 88 (vi) 98
(vii) 77 (viii) 96 (ix) 23 (x) 90
Solution:
(i) 729
729 3 3 3 3 3 3
729 3 3 3
27 (ii) 400
400 2 2 2 2 5 5
400 2 2 5 20.
(iii) 1764
1764 2 2 3 3 7 7
1764 2 3 7 42. (iv) 4096
4096 2 2 2 2 2 2 2 2 2 2 2 2
4096 2 2 2 2 2 2 64.
(v) 7744
7744 11 11 2 2 2 2 2 2
7744 11 2 2 2 88. (vi) 9604
9604 2 2 7 7 7 7
9604 2 7 7 98.
(vii) 5929
5929 7 7 11 11
5929 7 11 77. (viii) 9216
9216 2 2 2 2 2 2 2 2 2 2 3 3
9216 2 2 2 2 2 3 96.
(xi) 529
529 23 23 
529 23 (x) 8100
8100 3 3 3 3 2 2 5 5
8100 3 3 2 5 90.
Question 5
For each of the following numbers, find the smallest whole number by which it
should be multiplied so as to get a perfect square number.
Also, find the square
root of the square number so obtained.
(i) 252 (ii) 180 (iii) 1008
(iv) 2028 (v) 1458 (vi) 768
(i) 7; 42 (ii) 5; 30 (iii) 7, 84 (iv) 3; 78 (v) 2; 54 (vi) 3; 48
Solution:
(i)
252 2 2 3 3 7
Here, prime 7 does not occur in a pair.
252 is not a perfect square.
If we multiply the given number by 7, then
252 7 2 2 3 3 7 7 1764
Now, each prime occurs in a pair
252 7 1764 
is a perfect square.
Thus, the required smallest number is 7.
Also,
1764 2 3 7 42. (ii) Thus,
180 2 2 3 3 5
Here, prime 5 does not occur in a pair.
180 is not a perfect square.
If we multiply the given number by 5, then
1 80 5 900 2 2 3 3 5 5
Now, each prime occurs in a pair.
180 5, . ., 900ie
is a perfect square.
Thus, the required smallest number is 5.
Also,
900 2 3 5 30. (iii) 1008
Thus,
1008 2 2 2 2 3 3 7
Here, prime 7 does not occur in a pair.
1008 is not a perfect square.
If we multiply the given number by 7, then
1008 7 7056 2 2 2 2 3 3 7 7
Now, each prime occurs in a pair.
1008 7 . ., 7056ie
is a perfect square.
Thus, the required smallest number is 7.
Also,
7056 2 2 3 7 84.
(iv) 2028
Thus,
2028 2 2 3 13 13
Here, prime 3 does not occur in a pair.
2028 is not a perfect square.
If we multiply the given number by 3, then
2028 3 6084 2 2 3 3 13 13
Now, each prime occurs in a pair.
2028 3 . ., 6084ie
is a perfect square.
Thus, the required smallest number is 3.
Also,
6084 2 3 13 78 (v)
1458 2 3 3 3 3 3 3
Here, prime 2 does not occur in a pair.
1458 is not a perfect square.
If we multiply the given number by 2, then
1458 2 2916 2 2 3 3 3 3 3 3
Now, each prime occurs in a pair.
1 458 2, i.e, 2916
is a perfect square.
Thus, the required smallest number is 2.
Also,
2916 2 3 3 3 54. (vi)
768 2 2 2 2 2 2 2 2 3
Here, prime 3 does not occur in a pair.
768 is not a perfect square.
If we multiply the given number by 3, then
768 3 2304 2 2 2 2 2 2 2 2 3 3
Now, each prime occurs in a pair.
768 3, i.e, 2304
is a perfect square.
Thus, the required smallest number is 3.
Also,
2304 2 2 2 2 3 48.
Question 6
For each of the following numbers, find the smallest whole number by which it
should be divided so as to get a perfect square.
Find the square root of the square number so obtained.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645
(v) 2800 (vi) 1620
(i) 7; 6 (ii) 13; 15 (iii) 11; 6 (vi) 5; 23 (v) 7; 20 (vi) 5; 18
Solution:
(i) 252
252 2 2 3 3 7
If we divide 252 by 7, then
252 7 36 2 2 3 3 This is a perfect square.
The required smallest number is 7.
Also,
(ii) 2925
2925 3 3 5 5 13
If we divide 2925 by 13, then
2925 13 225 3 3 5 5
This is a perfect square.
The required smallest number is 13.
Also,
225 3 5 15. (iii) 396
396 2 2 3 3 11
If we divide 396 by 11, then
396 11 36 2 2 3 3
This is a perfect square.
The required smallest number is 11.
Also,
36 2 3 6
(iv) 2645
2645 5 23 23
If we divide 2645 by 5, then
2645 5 529 23 23
which is a perfect square.
The required smallest number is 5.
Also,
529 23 (v) 2800
2800 2 2 2 2 5 5 7
If we divide 2800 by 7, then
2800 7 400 2 2 2 2 5 5,
Which is a perfect square.
The required smallest number is 7.
Also,
400 2 2 5 20.
(vi) 1620
1620 2 2 3 3 3 3 5
If we divide 1620 by 5, then
1620 5 324 2 2 3 3 3 3
This is a perfect square.
The required smallest number is 5.
Also,
324 2 3 3 18. Question 7
The students of Class VIII of a school donated Rs. 2401 in all, for prime Minister’s
National Relief Fund.
Each student donated as many rupees as the number of students in the class.
Find the number of students in the class.
49
Solution:
2401 7 7 7 7
2401 7 7 49
Thus, the number of students in
Class VIII is 49.
Question 8
2025 plants are to be planted in a garden in such a way that each row contains as
Many plants as the number of rows.
Find the number of rows and the number of
Plants in each row.
45, 45
Solution:
Let the number of plants contained in each row be x.
Then, the number of rows will also be x.
According to the given condition of the question. 2
2025 2025x x x
2
2 2 2 2 2
5 3 3 5 3 3xx
22
45 45xx
Thus, the number of rows and the number of plants in each row is 45.
Question 9
Find the smallest square number that is divisible by each of the number 4, 9, and
10.
900
Solution:
This has to be done in three steps.
Step.1: First find the LCM of the numbers
4, 9 and 10
LCM of 4, 9 and
10 180
Step 2: Multiply the number 180 by 5 to make
it a perfect square.
180 5 2 2 3 3 5 5
900 2 2 3 3 5 5
Now, 900 is obviously a perfect square.
Thus, the required square number is 900.
180 2 2 3 3 5 Question 10
Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
3600
Solution:
LCM of 8, 15 and
20 120.
Prime factorization of
120 2 2 2 3 5
We see that 2, 3 and 5 are not in pairs.
Therefore, 120 should be multiplied by
2 3 5, . ., 30ie
in order to get a perfect square.
Hence, the required square number
120 30 3600. Exercise 6.4
Question 1
Find the square root of each of the following numbers by division method:
(i) 2304 (ii) 4489 (iii) 3481 (iv) 529
(v) 3249 (vi) 1369 (vii) 5776 (viii) 7921
(ix) 576 (x) 1024 (xi) 3136 (xii) 900
(i) 48 (ii) 67 (iii) 59 (iv) 23 (v) 57
(vi) 37 (vii) 76 (viii) 89 (ix) 24 (x) 32
(xi) 56 (xii) 30
Solution:
(i)
2304 48 (ii)
4489 67
(iii)
3481 59 (iv)
529 23
(v)
3249 57 (vi)
1369 37
(vii)
5776 76 (viii)
7921 89
(ix)
576 24 (x)
1024 32
(xi)
3136 56 (xii)
900 30
Question 2
Find the number of digits in the square root of the following numbers
(Without any calculation):
(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625
(i) 1 (ii) 2 (iii) 2 (iv) 3 (v) 3
Solution:
(i) 64: The number of digits in the square root is 1.
(ii) 144: The number of digits in the square root is 2.
(iii) 4489: The number of digits in the square root is 2.
(iv) 27225: The number of digits in the square root is 3.
(v) 390625: The number of digits in the square root is 3.
Question 3
Find the square root of the following decimal numbers:
(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
(i) 1.6 (ii) 2.7 (iii) 7.2 (iv) 6.5 (v) 5.6 Solution:
(i)
2.56 1.6
(ii)
7.29 2.7 (iii)
51.84 7.2
(iv)
42.25 6.5 (v)
31.36 5.6
Question 4
Find the least number which must be subtracted from each of the following numbers
so as to get a perfect square.
Also find the square root of the perfect square
so obtained.
(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000
(i) 2; 20 (ii) 53; 44 (iii) 1; 57 (iv) 41; 28 (v) 31; 63
Solution:
(i) Here, we get 2 as remainder. It shows that 20
2
is less than 402 by 2.
Thus, the number to be subtracted so as to make it a perfect square is 2.
Required number
402 2 400
And
400 20. (ii) Here, we get remainder 53. It shows that 44
2
is less than1989 by 53.
Thus, the number to be subtracted so as to make it a perfect square is 53.
Required number
1989 53 1936 
And
1936 44
\
(iii) Here, we get the remainder 1. It shows that 57
2
is less than 3250 by 1.
Thus, the number to be subtracted so as to make a perfect square is 1.
Required number
3250 1 3249
and
3249 57 (iv) Here, we get the remainder 41. It shows that 28
2
is less than825 by 41.
Thus, the number to be subtracted so as to make a perfect square is 41.
Required number
825 41 784
And
784 28
(v) Here, we get the remainder 31.
Thus, the smallest number to be
subtracted so as to make a perfect square is 31.
Required number
4000 31 3969
And
3969 63 Question 5
Find the least number which must be added to each of the following numbers so as
to get a perfect square.
Also, find the square root of the perfect square so obtained.
(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412
(i) 4; 23 (ii) 14; 42 (iii) 4; 16 (iv) 24; 43 (v) 149; 81
Solution:
(i) Remainder = 41
2
22 525
Next perfect square number
2
23 529.
Hence, the number to be added
529 525 4. 
The required number
525 4 529
And
529 23 (ii) Remainder = 69
Next perfect square number
2
42 1764.
Hence, the number to be added
1760 1750 14 
The required number
1750 14 1764
And
1764 42 (iii) Remainder = 27
2
15 252
Next perfect square number
2
16 256.
Hence, the number to be added
256 252 4.
The required number
252 4 256
And
256 16 (iv) Remainder = 61
2
42 1825
Next perfect square number
2
43 1849.
Hence, the number to be added
1849 1825 24.
The required number
1825 24 1849
And
1849 43 (v) Remainder = 12
2
80 6412 
Next perfect square number
2
81 6561. 
Hence, the number to be added
6561 6412 149 
The required number
6412 149 6561
And
6561 81
Question 6
Find the length of the side of a square whose area is 441 m
2
.
21m
Solution:
Let the side of the square be x m.
22
Area m x
i.e.,
2
441x
Or,
21x
Thus, the required length of square is 21 m. Question 7
In a right triangle
ABC, B 90 .
(i)
If AB = 6 cm, BC = 8 cm, find AC.
(ii)
If AC = 13 cm, BC = 5 cm, find AB.
i 10 cm ii 12 cm
Solution:
i Here, AB = 6 cm, BC = 8 cm.
Using Pythagoras theorem, we have:
2 2 2
AC = AB + BC
2 2 2
or,AC (36 64) cm 100 cm
Or,
AC = 10 cm.
(ii) Here,
AC = 13 cm, BC = 5 cm
Using Pythagoras theorem, we have:
2 2 2
AC = AB + BC
2 2 2 2 2
Or,AB AC BC (169 25) cm 144 cm
Or,
AB = 12 cm.
Question 8
A gardener has 1000 plants.
He wants to plant these in such a way that the number of rows and the number of
columns remain the same.
Find the minimum number of more plants he needs for this.
24
Solution:
1024 32 32
is a perfect square nearest to 1000
More number of plants he needs
1024 1000 24 Question 9
There are 500 children in a school.
For a P.T. drill they have to stand in such a
Manner that the number of rows is equal to the number of columns.
How many children would be left out in this arrangement?